Let y = x^2 e^{-x}. The interval in which y i | vedclass

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(D) Given the function $y = x^2 e^{-x}$.To find the interval where $y$ increases,we calculate the derivative $\frac{dy}{dx}$:$\frac{dy}{dx} = \frac{d}

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$( 0, 2 )$

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(D) Given the function $y = x^2 e^{-x}$.To find the interval where $y$ increases,we calculate the derivative $\frac{dy}{dx}$:$\frac{dy}{dx} = \frac{d}{dx}(x^2) \cdot e^{-x} + x^2 \cdot \frac{d}{dx}(e^{-x})$$\frac{dy}{dx} = 2x e^{-x} - x^2 e^{-x}$$\frac{dy}{dx} = x e^{-x} (2 - x)$For the function to be increasing,we require $\frac{dy}{dx} > 0$.Since $e^{-x}$ is always positive for all real $x$,the sign of $\frac{dy}{dx}$ depends on $x(2 - x)$.We need $x(2 - x) > 0$,which is equivalent to $x(x - 2) This inequality holds when $x$ lies between the roots $0$ and $2$.Therefore,the interval is $x \in (0, 2)$.

Let $y = x^2 e^{-x}$. The interval in which $y$ increases with respect to $x$ is:

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